Scalar to the power vector I have a formula that contains $\log (I + e^{X})$ term where $X$ is an $N$ dimensional vector ($x_0, x_1, ..., x_{N-1}$) and $I$ is the unity vector. I don't know how the exponentiation and log operations are defined in the case of vector. How to express this formula in terms of individual components of $X$?
Is $$e^X = [e^{x_0},e^{x_1},...,e^{x_{n-1}}]$$
and
$$\log(V) = [\log(v_0), \log(v_1), ..., \log(v_{N-1})]$$
?
 A: Almost all of the time, if you don't know what $e^X$ means in some context, it means the usual power series:
$$e^X = 1 + X + \frac{X^2}{2} + \ldots$$
Of course, this relies on a notion of a "product" of two vectors, and it's not automatically clear what that means. The obvious choice is the geometric product, but you'd know if it was that.
Your use of the phrase "identity vector" is suggestive. There is no such thing, so you may have meant something other than "vector". I think you might be referring to operators, i.e. matrices.
Edit
It turns out that all is well, and these are all scalars. The actual equation is of the form:
$$\mathbf{y} = C \log ( 1 + e^{C^{-1} \mathbf{x}} )$$
...where $C$ is a discrete Fourier transform. (It's actually in cepstral space rather than the usual frequency space, but anyone playing along at home can ignore this detail.)
The inverse DFT transforms the vector of frequency components back into a continuous function, and it is this function which goes through $\exp$ and $\log$. At the final step, the manipulated continuous function is re-transformed back into a frequency vector.
Did that make sense?
A: If $A$ is a matrix, you can use the Taylor formula: for example, $$\exp(A)=\sum_{k=0}^{\infty}\dfrac{A^k}{k!}$$ If $A$ is a $N$ dimensional vector you have: $f(A)=\left[f(A_1),f(A_2,...f(A_N)\right]$
